Problem: Simplify; express your answer in exponential form. Assume $k\neq 0, t\neq 0$. $\dfrac{{(k^{-2}t)^{3}}}{{kt^{2}}}$
To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(k^{-2}t)^{3} = (k^{-2})^{3}(t)^{3}}$ On the left, we have ${k^{-2}}$ to the exponent ${3}$ . Now ${-2 \times 3 = -6}$ , so ${(k^{-2})^{3} = k^{-6}}$ Apply the ideas above to simplify the equation. $\dfrac{{(k^{-2}t)^{3}}}{{kt^{2}}} = \dfrac{{k^{-6}t^{3}}}{{kt^{2}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{-6}t^{3}}}{{kt^{2}}} = \dfrac{{k^{-6}}}{{k}} \cdot \dfrac{{t^{3}}}{{t^{2}}} = k^{{-6} - {1}} \cdot t^{{3} - {2}} = k^{-7}t$